No. 49 - Longest Substring without Duplication

Problem: Given a string, please get the length of the longest substring which does not have duplicated characters. Supposing all characters in the string are in the range from ‘a’ to ‘z’.

Analysis: It’s not difficult to get all substrings of a string, and to check whether a substring has duplicated characters. The only concern about this brute-force strategy is performance. A string with n characters has O(n²) substrings, and it costs O(n) time to check whether a substring has duplication. Therefore, the overall cost is O(n³).

We may improve the efficiency with dynamic programming. Let’s denote the length of longest substring ending with the i^th character by L(i).

We scan the string one character after another. When the i^th character is scanned, L(i-1) is already know. If the i^th character has not appeared before, L(i) should be L(i-1)+1. It’s more complex when the i^th character is duplicated. Firstly we get the distance between the i^th character and its previous occurrence. If the distance is greater than L(i-1), the character is not in longest substring without duplication ending with the (i-1)^th character, so L(i) should also be L(i-1)+1. If the distance is less than L(i-1), L(i) is the distance, and it means between the two occurrence of the i^th character there are no other duplicated characters.

This solution can be implemented in Java as the following code:

public static int longestSubstringWithoutDuplication(String str) {

    int curLength = 0;

    int maxLength = 0;

    int position[] = new int[26];

    for(int i = 0; i < 26; ++i) {

        position[i] = -1;

    }

    for(int i = 0; i < str.length(); ++i) {

        int prevIndex = position[str.charAt(i) - 'a'];

        if(prevIndex < 0 || i - prevIndex > curLength) {

            ++curLength;

        }

        else {

            if(curLength > maxLength) {

                maxLength = curLength;

            }

            curLength = i - prevIndex;

        }

        position[str.charAt(i) - 'a'] = i;

    }

    if(curLength > maxLength) {

        maxLength = curLength;

    }

    return maxLength;

}

L(i) is implemented as curLength in the code above. An integer array is used to store the positions of each character.

Code with unit tests is shared at http://ideone.com/CmY3xN.

More coding interview questions are discussed in my book <Coding Interviews: Questions, Analysis & Solutions>. You may find the details of this book on Amazon.com, or Apress.

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